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I have a question about "old covariant quantization" in Polchinski's string theory p. 123.

It is said

The only nontrivial condition at this level is $(L_0^{\rm m} + A) | \psi \rangle =0 $, giving $m^2=A/\alpha'$.

I have no idea where does $m^2=A/\alpha'$ come from... How to get this result?

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1 Answer 1

up vote 3 down vote accepted

In the expression $(4.1.11a)$ :

$$L^m_0 = \alpha' p^2 + \alpha_{-1}.\alpha_1 + ....$$ You see that all the operators terms have a anihilation operator at the right (the $\alpha_n $)

So, applying $L^m_0$ to a ground state $|0;k\rangle$ gives zero for the operator part, so you have :

$$L^m_0|0;k\rangle =\alpha' p^2 |0;k\rangle = \alpha' k^2 |0;k\rangle = -\alpha' m^2 |0;k\rangle$$

[The Polchinski convention for the metrics is $g=(-1,+1,+1.....+1)$

Because we have a condition $(L^m_0 + A)|\psi \rangle =0$, we have :

$$0 = (L^m_0 + A)|0;k\rangle = (A - \alpha' m^2 )|0;k\rangle$$

So, finally :

$$A = \alpha' m^2 $$

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